Fundamentals of Plasma Physics

4.4 Two-fluid model of Alfvén modes 141

when the perpendicular current, i.e.,J 1 ⊥=

∑

nqσuσ⊥ 1 is considered, the electron and ion
E×Bdrift terms cancel so that the polarization terms become the leading terms involving
the electric field. Because of the mass in the numerator, the ion polarization drift is much
larger than the electron polarization drift. Thus, the perpendicular current comes from ion
polarization drift and diamagnetic current

μ 0 J⊥ 1 =

μ 0 nmiE ̇⊥

B^2

−

∑

σ

∇Pσ⊥ 1 ×B

B^2

=

1

vA^2

E ̇⊥ 1 −μ^0 ∇P⊥^1 ×B

B^2

(4.99)

whereP⊥ 1 =

∑

Pσ⊥ 1 .The term involvingP ̇⊥ 1 has been dropped because it is small by
ω/ωccompared to theP⊥ 1 term.
The center of mass perpendicular motion is

U⊥ 1 =

∑

mσnuσ⊥ 1

∑

mσn

≈ui⊥ 1 (4.100)

An important issue for the perpendicular motion is whetheruσ⊥ 1 is compressible or incom-
pressible. Let us temporarily ignore parallel motion and consider the continuity equation

∂n 1

∂t

+n∇·uσ⊥ 1 =0. (4.101)

If∇·uσ⊥ 1 =0,the mode does not involve any density perturbation, i.e.,n 1 =0, and
is said to be an incompressible mode. On the other hand, if∇·uσ⊥ 1 =0then there are
fluctuations in density and the mode is said to be compressible.
To proceed further, consider the vector identity

∇·(F×G)=G·∇×F−F·∇×G.

IfGis spatially uniform, this identity reduces to∇·(F×G)=G·∇×Fwhich in turn
vanishes ifFis the gradient of a scalar (since the curl of a gradient is always zero). Taking
the divergence of Eq.(4.98) and ignoring the polarization terms (they are of orderω/ωci
and are only important when calculating the current which we are not interested in right
now) gives

∇·uσ⊥ 1 =

1

B^2

B·∇×E 1 =

1

B

zˆ·∇×E 1 (4.102)

to lowest order. SettingE 1 =−∇φ(i.e., assuming that the electric field is electrostatic)
would cause the right hand side of Eq.(4.102) to vanish, but such an assumption is overly
restrictive because all that matters here is thez-component of∇×E 1 .Thez-component
of∇×E 1 involves only the perpendicular component of the electric field (i.e., onlythex
andycomponents of the electric field) and so the least restrictive assumption for the right
hand side of Eq.(4.102) to vanish is to haveE 1 ⊥=−∇⊥φ.Thus, one possibility is to have
E 1 ⊥=−∇⊥φin which case the perpendicular electric field is electrostatic in nature and
the mode is incompressible.
The other possibility is to haveˆz·∇×E 1 =0. In this case, invoking Faraday’s law
reduces Eq.(4.102) to

∇·uσ⊥ 1 = −

1

B

ˆz·

∂B 1

∂t

= −

1

B

∂Bz 1

∂t

. (4.103)